Difference in liquid column heights in a rotating U-tube

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Homework Help Overview

The discussion revolves around a U-shaped tube containing a liquid, which is mounted on a rotating turntable. Participants are exploring the difference in height between the liquid columns in the vertical arms due to the effects of rotation, specifically focusing on the implications of angular speed on pressure differences within the tube.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between linear acceleration and angular speed, questioning whether to use average acceleration in their calculations. There is also exploration of pressure gradients and how they relate to the liquid column heights.

Discussion Status

Some participants have provided insights into the pressure differences and have attempted to derive equations related to the problem. However, there remains uncertainty regarding the integration process and the application of concepts from linear acceleration to the circular case.

Contextual Notes

Participants note potential confusion regarding the integration of pressure differences and the concept of pressure gradients. There is also mention of previous parts of the problem that may influence their current understanding.

kidsmoker
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Homework Statement



A U-shaped tube with a horizontal segment of length L contains a liquid. What is the difference in height between the liquid columns in the vertical arms if the tube is mounted on a horizontal turntable, and is rotating with angular speed w, with one of the vertical arms on the axis of rotation?

Homework Equations



a=w^2r towards the centre
p = p(o) + density*g*h

The Attempt at a Solution



A previous part to the question was finding the difference in height if the U-tube had acceleration a towards the right. I did this and found h=aL/g.

I was thinking I could just substitute a=w^2L into this equation but then I had second thoughts since the acceleration decreases as you get closer to the axis of rotation. Do you take the average acceleration intead?

Thanks.
 
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kidsmoker said:
I was thinking I could just substitute a=w^2L into this equation but then I had second thoughts since the acceleration decreases as you get closer to the axis of rotation. Do you take the average acceleration intead?

Hi kidsmoker! :smile:

In the first case (linear acceleration),

if the linear density (mass/length) is ρ, and the pressure gradient is Q, then a length dr in the middle will have mass ρ dr and acceleration a, and so mass x acceleration = aρ dr = pressure difference = Q dr,

so Q = aρ, and so total pressure difference = aρL.

Does that help you with the circular case? :smile:
 
Ah yeah I get it now thanks. :)
 
Although I took vector calculus, I didn't understand what you mean by "gradient pressure", tiny-tim. I think I should google for it.

For the first question I get that the answer is h=\frac{aL}{g} and the column of liquid that has a higher high is the right one.

I tried the second part of the question, namely the first question that kidsmoker asked, and reached h=\frac{L^2 \omega ^2}{2g} but I'm unsure of the integral I used.
The difference of pressure within a dr element is \omega ^2 r \rho dr. I integrated this expression with respect to r, from 0 to L.
 

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